Test: Limits And Derivatives (CBSE Level) - 1


25 Questions MCQ Test Mathematics For JEE | Test: Limits And Derivatives (CBSE Level) - 1


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QUESTION: 1

 is equal to 

Solution:

We have:
limx→0 ln(1+ax)]/x=a
lim x→0 ln(1+ax)/ax=a
and then:
lim x→0 [ln(1+ax)-ln(1+bx)]/x
= a-b

QUESTION: 2

 is equal to

Solution:

On rationalizing,
 Lt x-->0 (1-Cosx)/[√(1+x) +1]
= lt x-->0 2sin2(x/2)[√(1+x)+1] /x
= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]
= 2.1/2 .0[1+1]=0

QUESTION: 3

Let f (x) = x sin 1/x, x ≠ 0, then the value of the function at x = 0, so that f is continuous at x = 0, is

Solution:

f(0) = lim(x→0) x sin (1/x)
​We know ∀ x ∈ R,sin (1/x) ∈ [−1,1]
Hence, f(0) = lim(x→0) xsin(1/x) 
= 0

QUESTION: 4

 is equal to 

Solution:

lim(x→0) (tanx−sinx)/x3
=lim(x→0) [sinx(x−cosx)]/x3cosx
=lim(x→0) (sinx)/x × (1−cosx)/x2 × (1/cosx)
= 1 × 1/2 × 1/1
= 1/2

QUESTION: 5

 is equal to 

Solution:

lim x→0 (1−cosx)/(xlog(1+x))
= lim x→0 {2sin2 x/2}/[xlog(1+x)]
= lim x→0 {{2sin2x/2/(x/2)2}*/1/4}/[1/xlog(1+x)]
= lim x→0 [log(1+x)/x]1/2
= 2​

QUESTION: 6

 is equal to

Solution:
QUESTION: 7

Solution:

Consider the given function.
lim x→0 ((1-x)n −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [n(1-x)n−1 −0)]/1
lim x→ 0(n(1-x)n−1)
=n(1-0)n−1
= n - 1

QUESTION: 8

 is equal to

Solution:

lim(x → 0) (tanx-x)/x2 tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3

QUESTION: 9

Solution:

 lim x-->0 ((x2cosx-)/1-cosx)
Differentiating numerator and denominator
lim x-->0 (2xcosx - x2sinx)/sinx
Differentiating it for a second time
lim x-->0 (2cos - 2xsinx - 2xsinx - x2cosx)/cosx
Substituting x = 0
= lim x-->0  2/1
= 2

QUESTION: 10

 is equal to 

Solution:

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x3.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)3,
= lim y→0 (3siny − 4sin3y)− 3y)/27y3,
= lim y→0 {3(siny − y)/27y3) − (4sin3y/27y3)},
⇒ L = lim y→0 1/9 * (siny − y)/y3) − 4/27* (siny/y)3...(∗)
.Note that, here,
= lim y→0(siny − y)/y3)
= lim x→0 (sinx−x)/x3) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

QUESTION: 11

 is equal to 

Solution:
QUESTION: 12

 then for f to be continuous at x = 0, f (0) must be equal to

Solution:
QUESTION: 13

Solution:
QUESTION: 14

 where a > 0, is equal to 

Solution:
QUESTION: 15

 is equal to 

Solution:
QUESTION: 16

 is equal to

Solution:
QUESTION: 17

is equal to

Solution:

lim(h → 0) [sin2(x+h) - sin2x]/h
l = lim(h→0) = [2sin(x+h)cos(x+h)]/1
{As we know that sin2x = 2sinxcosx
= sin2x

QUESTION: 18

If a is a real number, then  

Solution:
QUESTION: 19

If k be a integer, then  is equal to

Solution:

- lim x-->k+ = x-{x}
Lim x→k+   k-{k}
                  = 0           

QUESTION: 20

is equal to

Solution:

Correct Answer :- C

Explanation : lt(x->a-) -|x-a|/x-a

= - |-a-a|/(-a-a)

= -|-2a|/(-2a)

= -2a/(-2a)

= 1

QUESTION: 21

 is equal to 

Solution:
QUESTION: 22

Maximum value of x3−3x+2in [0,2] is

Solution:
QUESTION: 23

Solution:

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

QUESTION: 24

 is equal to 

Solution:
QUESTION: 25

The value of the limit Limx→0 (cos x)cot2 x is

Solution:

Given, Limx→0 (cos x)cot² x
= Limx→0 (1 + cos x – 1)cot² x
= eLimx→0 (cos x – 1) × cot² x
= eLimx→0 (cos x – 1) / tan² x
= e-1/2